Answer:
P.E. = -0.449 J
Explanation:
Potential energy of a charge particle in any electrostatic field is defined as the amount of work done ( in negative ) to bring that charge particle from any position to a new position r.
Now Potential energy is defined by this formula,
P.E. = k q₁ q₂/ r
where P.E. is the potential energy.
k = 1/( 4πε₀) = 8.99 × 10⁹ C²/ ( Nm²)
q₁ = charge of one particle = +1.0μC
q₂ = charge of another particle = -5.0μC
r = distance = 0.1 m
Now , P.E. = 8.99 × 10⁹C²/ ( Nm²) * ( -5.0 × 10⁻⁶ C ) × ( 1 × 10⁻⁶ C ) / 0.1 m
P.E. = -0.449 J
Answer: The spring constant is K=392.4N/m
Explanation:
According to hook's law the applied force F will be directly proportional to the extension e produced provided the spring is not distorted
The force F=ke
Where k=spring constant
e= Extention produced
h=2m
Given that
e=20cm to meter 20/100= 0.2m
m=100g to kg m=100/1000= 0.1kg
But F=mg
Ignoring air resistance
assuming g=9.81m/s²
Since the compression causes the plastic ball to poses potential energy hence energy stored in the spring
E=1/2ke²=mgh
Substituting our values to find k
First we make k subject of formula
k=2mgh/e²
k=2*0.1*9.81*2/0.1²
K=3.921/0.01
K=392.4N/m
Explanation:
It is given that,
Length of wire, l = 0.53 m
Current, I = 0.2 A
(1.) Approximate formula:
We need to find the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire, r = 2 cm = 0.02 m
The formula for magnetic field at some distance from the wire is given by :
B = 0.000002 T
(2) Exact formula:
B = 0.00000199 T
or
B = 0.000002 T
Hence, this is the required solution.
